Mathematica
Balkanica
––––––––––
New Series Vol. 16, 2002, Fasc. 1-4
Pure Multigrades
A.W. Goodman
This article is dedicated to the 70th anniversary of Acad. Bl. Sendov
(1)
Two sets of integers A and B form a pure multigrade of type (D, n) if A ∩ B is empty,
X d
X d
xk =
yk
d = 0, 1, 2, . . . , D,
xk ∈A
yk ∈B
and |A| = |B| = n. A pure multigrade is perfect, if A ∪ B = {1, 2, . . . , 2n}. We prove some
new theorems about pure multigrades and state some open problems.
AMS Subj. Classification: Primary 11A99, 11D04; Secondary 11P99, 11Y99
Key Words: multigrade, pure multigrade, perfect multigrade, primary multigrade
1. Introduction
A multigrade of type (D, n) is a pair of multisets A and B of integers
such that |A| = |B| = n, and
X
X
(2)
xdk =
ykd
xk ∈A
yk ∈B
for d = 1, 2, . . . , D. Here, D is the degree of the multigrade (A, B) and n is the
size.
The word multiset implies that some integers may occur more than once
in A or B. Equation (2) and its various generalizations and specializations have
been investigated for many years; see Gloden [1] and Borwein and Ingalls [2].
Clearly, if A and B contain a common element q it can be dropped from
both sets without harm to the condition (2). In this paper, A and B will be
true sets with no repetitions permitted.
We say that A and B form a pure multigrade (A, B) of type (D, n) if
A and B are sets that satisfy equation (2) and A ∩ B = ∅. Equation (2) can
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A.W. Goodman
be extended to include more than two sets and in this situation the existence
of a solution was widely known as the Tarry-Escott problem. It was a surprise
when Wright [3] pointed out that Prouhet [4] had announced that the enlarged
system ((2) for more than two sets) always has a solution and indicated his
method for computing such asolution. Further, Wright explained in some detail
the solution that Prouhet had merely hinted at. Prouhet’s work was published
70 years before the Tarry and Escott papers appeared and 70 years before the
problem was formulated. However, there are still many open questions related
to equation (2) and its variations.
We will need two well-known theorems that are basic to the subject.
We say that the set S = {u1 , u2 , . . . , un } is subjected to the affine transformation T : v = au + b, where a and b are integers if T (S) is the set
{vk = auk + b | k = 1, 2, . . . , n}. Then we have the following theorem.
Theorem 1.
(Frolov [5]) If the sets A and B satisfy the system of
equations (2), then for any T the sets T (A) and T (B) satisfy the same system
of equations.
Thus, the transformation T takes a multigrade of type (D, n) into another
multigrade of the sametype.
Theorem 1 shows that for a multigrade (A, B) the sets A or B may
contain negativeintegers, so some normalization should be made if we wish to
avoid this.
We will call a multigrade, (A, B), normalized if 1 ∈ A and all other
integers in A ∪ B are greater than one. It is clear that any pure multigrade of
type (D, n) can be transformed into a normalized pure multigrade of the same
type by using a suitable T and Theorem 1. Gloden [1, p. 21] has a different
definition for a normalized multigrade.
Two multigrades, (A, B) and (C, D), are said to be equivalent if either
can be obtained from the other by a suitable affine transformation T : v = au+b,
where a and b are integers. Notice that this set of transformations does not form
a group, but only a semigroup, because the inverse of atransformation may not
have the proper form.
Theorem 2. (Bastien [6]) If (A, B) is a pure multigrade of type (D, n),
then
(3)
n ≥ D + 1.
Pure Multigrades
65
It is convenient to have the standard notation
D
x1 , x2 , . . . , xn = y1 , y2 , . . . , yn
for a solution of (2). With this notation we have the following examples of pure
normalized multigrades of type (2, n):
2
(4)
n=3
1, 5, 6 = 2, 3, 7
(5)
n=4
1, 4, 6, 7 = 2, 3, 5, 8
(6)
n=5
1, 5, 8, 10, 13 = 2, 3, 9, 11, 12
2
2
2. The union of two multigrades
If (A1 , B1 ) and (A2 , B2 ) are two multigrades of type (D1 , n1 ) and (D2 , n2 ),
respectively, we can form a new multigrade (A∗ , B ∗ ) by taking their union
(7)
A∗ = A1 ∪ A2
and B ∗ = B1 ∪ B2 .
We indicate this union by writing
(A∗ , B ∗ ) = (A1 , B1 ) ∪ (A2 , B2 ) .
Here (A∗ , B ∗ ) has type (D∗ , n∗ ), where n∗ = n1 + n2 and D∗ = min {D1 , D2 }.
Henceforth, we will assume that D1 = D2 = D.
If (A1 , B1 ) and (A2 , B2 ) are pure multigrades, it does not follow that
∗
∗
(A , B ) will be a pure multigrade. But the union will be a pure multigrade of
type (D, n∗ ) if max {A1 ∪ B1 } < min {A2 ∪ B2 }. For example, if we transform
2
the multigrade (5) by T : v = u+7, then we obtain the multigrade 8, 11, 13, 14 =
9, 10, 12, 15. Then the union of this multigrade and the one given by (4) is:
(8)
2
1, 5, 6, 8, 11, 13, 14 = 2, 3, 7, 9, 10, 12, 15,
a normalized multigrade of type (2, 7).
Definition 1. A pure multigrade is said to be primitive, if it cannot
be obtained as the union of two pure multigrades with no elements in common.
At present, we do not have any efficient general method for determining
whether a pure multigrade is primitive or not. However, we can prove the
following.
Theorem 3. If D ≥ 2 and n ≤ 5, then a pure multigrade of type (D, n)
is primitive.
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A.W. Goodman
P r o o f. If the components in the union have type (D, n1 ) and (D, n2 ),
then n1 + n2 ≤ 5. Hence nk ≤ 2 for some k, and by Bastien’s theorem this is
impossible, if D ≥ 2.
Theorem 4. The decomposition of a pure multigrade into the union of
primitive pure multigrades is not always unique.
P r o o f. We consider the example
(9)
2
1, 4, 6, 7, 9, 12, 15, 19, 24 = 2, 3, 5, 8, 10, 11, 14, 21, 23 ,
a pure multigrade of type (2, 9). A reasonable amount of computation will show
that
(10)
(A, B) = (C1 , D1 ) ∪ (E1 , F1 )
and (A, B) = (C2 , D2 ) ∪ (E2 , F2 ) ,
where
(11)
C1 = {1, 4, 6, 7}
E1 = {9, 12, 15, 19, 24}
D1 = {2, 3, 5, 8}
F1 = {10, 11, 14, 21, 23}
and
(12)
C2 = {5, 8, 10, 11}
E2 = {2, 3, 14, 21, 23}
D2 = {6, 7, 9, 12}
F2 = {1, 4, 15, 19, 24} .
Now each of the sets in (11) and (12) has n = 4 or 5, so each of the multigrades in
the unions (10) is a primitive multigrade. The multigrade (C1 , D1 ) isequivalent
to (C2 , D2 ) under T : v = u + 4 but (E1 , F1 ) is not equivalent to either (E2 , F2 )
or (F2 , E2 ). Hence, (10) gives a decomposition of (A, B) into a union of primitive
multigrades in two different ways.
Counting the numbers of primes less than a given constant has long
been a fascinating problem. Thus, it may be of interest to count the number
of primitive normalized pure multigrades for which M = max{A ∪ B} ≤ M0 .
Here, of course, if a number of primitive multigrades are equivalent we count
only the one with the smallest M .
3. Perfect multigrades
It will be very convenient to have a short symbol for certain sums that
occur frequently. Thus we will let
X
Sd (C) =
(13)
cdk .
ck ∈C
Pure Multigrades
67
We will call a pure multigrade of type (D, n) perfect, if
A ∪ B = {1, 2, . . . , 2n},
the first 2n integers.
(2, n).
Theorem 5. If n = 4k, with k ≥ 1 there is a perfect multigrade of type
P r o o f. If k = 1 we have the perfect multigrade given in equation (5),
where A ∪ B = {1, 2, 3, 4, 5, 6, 7, 8}.
For larger values of k we first observe that for any integer q, we have
q + (q + 3) = (q + 1) + (q + 2) so we can place one pair in A and one pair
in B without disturbing the equality ΣA x = ΣB y. For the squares we have
q 2 + (q + 3)2 − (q + 1)2 − (q + 2)2 = 4, so if one pair is in A and the other pair
is in B, the equality ΣA x2 = ΣB y 2 is unbalanced by 4. But if |A| = |B| = 4k,
we can alternate the placement of these pairs so that ΣA x2 = ΣB y 2 .
For example, if k = 3 so that 4k = 12 and |A ∪ B| = 24, we can set
A = {1, 4, 6, 7, 9, 12, 14, 15, 17, 20, 22, 23}
and
B = {2, 3, 5, 8, 10, 11, 13, 16 18, 19, 21, 24}.
Then (A, B) is a perfect normalized multigrade of type (2, 12).
In general, if n = 4k, this method will give a perfect normalized multigrade of type (2, 4k) in which M = max{A ∪ B} = 8k = 2n.
Theorem 6. If n = 4k + 2 with k ≥ 1, there is a perfect multigrade of
type (2, n).
P r o o f. For k = 1 we have n = 6. Set A = {1, 3, 7, 8, 9, 11} and B =
{2, 4, 5, 6, 10, 12}. Then S1 (A) = S1 (B) = 39 and S2 (A) = S2 (B) = 325, so
(A, B) form a perfect multigrade of type (2, 6). For larger values of k we add
integers q, q + 3, q + 1, q + 2 in alternating pairs as in the proof of Theorem 5.
For example if k = 2, so n = 10 we have
A = {1, 3, 7, 8, 9, 11, 13, 16, 18, 19}
and
B = {2, 4, 5, 6, 10, 12, 14, 15, 17, 20} .
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A.W. Goodman
Here S1 (A) = S1 (B) = 105 and S2 (A) = S2 (B) = 1435.
2n
P
If n is odd, then k = n(2n + 1), an odd number. Thus a partition with
1
S1 (A) = S1 (B) is impossible. Consequently, there are no perfect multigrades
with n odd. But if n is odd, there are multigrades that are as close to perfect
as possible, namely A ∪ B ⊂ {1, 2, 3, . . . , 2n + 1}.
Theorem 7. If n = 4k + 1 and k ≥ 2, then there is a normalized pure
multigrade of type (2, n) with M = max(A ∪ B) = 2n + 1.
P r o o f. For k = 2 we have n = 9 and we set
A = {1, 3, 7, 8, 10, 11, 13, 17, 19},
and
B = {2, 4, 5, 6, 9, 14, 15, 16, 18}.
Here S1 (A) = S1 (B) = 89 and S2 (A) = S2 (B) = 1163.
For larger values of k add integers q, q + 3, q + 1, q + 2 in alternating
pairs as in the proof of Theorem 5.
If k = 1, there is NO normalized pure multigrade of type (2, 5) with
M = 11.
Theorem 8. If n = 4k + 3 and k ≥ 0, then there is a normalized pure
multigrade of type (2, n) with M = 2n + 1.
P r o o f. For k = 0, use the multigrade given in equation (4), where
M = 7 = 2(3) + 1.
For larger values of k add integers q, q + 3, q + 1, q + 2 in alternating
pairs as in the proof of Theorem 5.
For example if k = 2, so n = 11 we have
A = {1, 5, 6, 8, 11, 13, 14, 16, 19, 21, 22}
and
B = {2, 3, 7, 9, 10, 12, 15, 17, 18, 20, 23}.
Here S1 (A) = S1 (B) = 136 and S2 (A) = S2 (B) = 2154.
Pure Multigrades
69
4. Multigrades as vectors
In this section, we regard any set C = {c1 , c2 , . . . , cn } as a vector with
components ci . As usual, addition of two such sets will be termwise and we will
be using the standard dot product A · B of two sets.
Theorem 9. Let (A, B) and (C, D) be two multigrades each of type
(2, n). Then the vector sums (A + C, B + D) will be a multigrade of the same
type iff as a dot product A · C = B · D.
n
n
P
P
ai + ci =
bi + di is trivial. For the sum of
P r o o f. The condition
1
1
the squares we need
n
X
(14)
(ai + ci )2 =
i=1
n
X
(bi + di )2 .
i=1
Since S2 (A) = S2 (B) and S2 (C) = S2 (D), then (14) holds iff
n
X
(15)
ai ci =
i=1
n
X
bi di ,
i=1
or A · C = B · D.
Of course the new multigrade will not always be perfect, or even pure,
but (A + C, B + D) will always have only integers and can always be normalized.
The following two examples show that the theorem is not empty.
Set A = {1, 5, 6}, B = {2, 3, 7}, C = {1, 20, 12}, and D = {10, 2, 21}.
Then (A, B) and (C, D) are multigrades of type (2, 3). Further, A · C = B ·
D = 173. Consequently, (A + C, B + D) is a multigrade of type (2, 3). Here,
(A+C) = {2, 25, 18} and (B+D) = {12, 5, 28} with S1 (A+C) = S1 (B+D) = 45
and S2 (A + C) = S2 (B + D) = 953.
For a more general example, we can use the formula given by Gloden [1,
p.33]. For any 8 integers a, b, c, d, r, s, t, and u let
A = {ab, cd, ac + bd},
B = {ac, bd, ab + cd},
C = {rs, tu, rt + su},
D = {rt, su, rs + tu}.
and
Then (A, B) and (C, D) are multigrades of type (2, 3). A brief computation will
show that A · C = B · D iff au = dr or cs = bt. It is clear that one can select
the variables in many ways so that all of the multigrades involved are pure.
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A.W. Goodman
5. Open problems
Let G be a collection of pure normalized multigrades (A, B) of type (2, 3)
with the following properties:
(1) no two multigrades in G are equivalent,
(2) every pure multigrade of type (2, 3) is equivalent to one of the multigrades
in G.
It is clear that such a set G exists. What is lacking, is a formula that will
give G. The beautiful formula given by Gloden [1, p.33] gives all multigrades
of type (2, 3), but it gives far too many multigrades. It gives some multigrades
that are NOT pure, it violates both the conditions for G, and there is no hint
of the restrictions on the parameters that will give a formula for G.
and
The sets (A, B) defined by
ª
©
A = 1, 2 + x + 2s + kx + sx, 2 + 2x + x2 + s − k + 2sx
ª
©
B = 2 + x + s − k, 1 + kx + sx, 2 + 2x + x2 + 2s + 2sx
satisfy the equations S1 (A) = S1 (B) and S2 (A) = S2 (B) for all values of the
three parameters k, s, and x. Hence, these formulas generate a multigrade of
type (2, 3) for all integervalues of the three parameters. What restrictions on
these parameters will turn this collection of pairs (A, B) into a set G?
We can introduce a measure L(D, n), called the length. As usual let
M = max{A ∪ B} and for fixed (D, n), let L(D, n) = min{M } for the set of all
normalized pure multigrades of type (D, n). For D = 2, the value of L(D, n)
was obtained in Theorems 5, 6, 7, and 8. For D ≥ 3, the value of L(D, n) is
unknown, but a computer search will give these values for small D and n.
During the preparation of this work, Prof. Gerald Myerson made important contributions which I am happy to acknowledge. Prof. Myerson proved
Theorems 6 and 7. He also found the nice conditions au = dr or cs = bt for
A · C = B · D in Theorem 9. Further, he found the two interesting examples
given at the end of that theorem.
In a letter, [8]
proved that for every D ≥ 2, there is a perfect
¢
¡ Myerson
multigrade of type D, 2D .
Pure Multigrades
71
References
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[3] E. M. W r i g h t. Prouhet’s 1851 solution of the Tarry-Escott Problem of
1910, Amer. Math. Monthly, 66, 1959, 199-201.
[4] E. P r o u h e t. Mémoire sur quelques relations entre les puissances des nombres, C. R. Acad. Sci., Paris, 33, 1851, p. 225.
[5] M. F r o l o v. Ègalités à deux degrés, Bulletin de la Société Mathematique
de France, 17, 1888-1889, 69-83.
[6] L. B a s t i e n. Sphinx-Oedipe 8, 1913, 171-172.
[7] R. M a l t b y. Pure product polynomials and the Prouhet-Tarry-Escott
problem, Math. Comp., 66, no. 219, 1997, 1323-1340.
[8] G. M y e r s o n. To appear.
Department of Mahematics
University of South Florida
Tampa, FL 33620-5700, USA
Author’s Home Address:
11321 Carrollwood Drive, Tampa, FL 33618, USA
e-mail: [email protected]
Received: 07.12.2001